Solutions. 15.1 Solubility Solution: homogeneous mixture or mixture in which components are...

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Solutions

15.1 SolubilitySolution: homogeneous mixture or mixture

in which components are uniformly intermingled

Solute: substance that is being dissolved in solvent

Solvent: substance that dissolves solvent and present in a large amount

Aqueous solutions: solutions with water as the solvent

Table 15.1 – Various Types of Solutions

15.2 Solution Composition: An IntroductionSaturated: A solution in which the

maximum amount of solvent has been dissolved. Any more solute added will sit as crystals on the bottom of the container

Unsaturated: A solution in which more of solute can be dissolved

Concentrated: a relative large amount of solute is being dissolved in solvent

Diluted: a relative small amount of solute is being dissolved in solvent

15.3 Solution Composition: Mass Percent

mass of soluteMass percent (m/m%) = ------------------------

x 100 mass of solution

ExamplesA solution is prepared by mixing 1.00 g of

ethanol, C2H5OH, with 100.0 g water. Calculate the mass percent of ethanol in this solution

A 135 g sample of seawater is evaporated to dryness, leaving 4.73 g of solid residue (the salts formerly dissolved in the seawater). Calculate the mass percent of solute present in the original seawater

Example: Determine Mass of SoluteAlthough milk is not a true solution (it is

really a suspension of tiny globules of fat, protein, and other substrates in water), it does contain a dissolved sugar called lactose. Cow’s milk typically contains 4.5 % by mass of lactose, C12H22O11. Calculate the mass of lactose present in 175 g of milk

15.4 Solution CompositionMolarity: the number moles of solute per

volume of solution in litersmoles of solute

Molarity = ------------------------Liters of solution

unit = moles/L or M (molar)

Standard solution: is a solution whose concentration is accurately known.

ExamplesCalculate the molarity of a solution prepared

by dissolving 11.5 g of solid NaOH in enough water to make 1.50 L of solution

Calculate the molarity of a solution prepared by dissolving 1.56 g of gaseous HCl into enough water to make 26.8 mL of solution

Determine how much volume (in ml) will be needed to dissolved 2.50 g of solid NaCl to make 0.050M solution.

E.g Solution Composition: Calculating Ion ConcentrationGive the concentration of all the ions in each

of the following solutions:0.50 M Co(NO3)2

1.0 M FeCl3

E.g Solution Composition: Calculating Number moles from MolarityHow many moles of Ag+ ions are present in

25.0 mL of a 0.75 M AgNO3 solution?

How many moles of Na+ ions are present in 42.0 mL of 0.350M NaCl?

Examples: Calculating mass from molarityTo analyze the alcohol content of a certain

wine, a chemist needs 1.00L of an aqeuous 0.200 M K2Cr2O7 (potassium dichromate) solution. How much solid K2Cr2O7 (molar mass = 294.2 g) must be weighed out to make this solution?

Formalin is an aqueous solutions of formaldehyde, HCHO,, used as a preservative for biological speciments. How many grams of formaldehyde must be used to prepare 2.5 L of 12.3 M formalin?

15.5 DilutionReducing the original concentration of a

chemical solutionA process of transferring solution to achieve

a the desired molarity by diluting with solvent

Moles of solute after dilution = moles of solute before dilution

Formula M1 V1 = M2 V2

ExamplesWhat volume of 16 M sulfuric acid must be

used to prepare 1.5L of 0.10 M of H2SO4 solution?

Calculate the new molarity if a dilution is made for:25.0 ml of water is added to 10.0 mL of 0.251

M CaCl2

15.6 Stoichiometry of Solution ReactionsSteps for solving stoichiometric problems

involving solutionsStep 1: Write a balanced equation for the

reaction. For each reactions involving ions, it is best to write the net ionic equation.

Step 2: Calculate the moles of reactantStep 3: Determine which reactant is limitingStep 4: Calculate the moles of other reactants

or products, as requiredStep 5: Convert to grams or other units, if

required

ExamplesWhen Ba(NO3)2 and K2CrO4 react in aqueous

solution, the yellow solid BaCrO4 is formed. Calculate the mass of BaCrO4 that forms when 3.50 x 10-3 mole of solid Ba(NO3)2 is dissolved in 265 mL of 0.0100 M K2CrO4 solution

ExamplesWhen aqueous solutions of Na2SO4 and

Pb(NO3)2 are mixed, PbSO4 precipitates. Calculate the mass of PbSO4 formed when 1.25 L of 0.0500 M Pb(NO3)2 and 2.00 L of 0.0250 M Na2SO4 are mixed

Neutralization ReactionUse stoichiometry to determine how much of

acid or base must be used to reach neutralization

Strong acid: HCl(aq) H+(aq) + Cl-(aq)

Strong base: NaOH(s) Na+(aq) + -OH(aq)

Net equation: H+(aq) + -OH(aq) H2O(l)

ExampleWhat volume of a 0.100 M HCl solution is

needed to neutralize 25.0 mL of a 0.350 M NaOH?

Calculate the volume of 0.10 M HNO3 needed to neutralize 125 mL of 0.050 MKOH

15.8 Solution Composition: NormalityNormality is another unit of concentration

sometime used when dealing with acid and base

H+ and –OHEquivalent of an acid: the mount of acid that

can be furnish 1 mol of H+ ionsEquivalent of a base: the amount of that base

that can furnish 1 mol of –OH ions

EquivalentE.g 1 mol HCl = 1 equiv HCl molar mass (HCl) = equivalent weight

(HCl)1 mol KOH = 1 equiv KOHmolar mass KOH = 1 equiv KOH½ mol H2SO4 = 1 equiv H2SO4

½ molar mass H2SO4 = 1 equiv H2SO4

Solution Stoichiometry: Calculating Equivalent WeightPhosphoric acid, H3PO4 can furnish three H+

ions per molecule. Calculate the equivalent weight of H3PO4.

Calculate the equivalent weight of HBr

Normality (N)Normality (N) = number of equivalent of

solute per liter of solution

Knowing Normality can help us calculateThe number of equivalentsThe volume of solution

Calculating NormalityA solution of sulfuric acid contains 86 g of

H2SO4 per liter of solution. Calculate the normality of this solution

Calculate the normality of a solution containing 23.6 g of KOH in 755 ml of solution

Neutralization The number of H+ ions furnished by the

sample of acid is the same as the number of –

OH ions furnished by the sample of base reacts exactly with

n equiv acid --------------- n equiv baseNacid x Vacid = Nbase x Vbase